Problem 34 - Precalculus Mathematics for Calculus

Question

A description of a line is given. Find parametric equations for the line. The plane that contains the line \(| x=1-t, y=2+t\) \(z=-3 t\) and the point \(P(2,0,-6) .\) [Hint: A vector from any point on the line to \(P\) will lie in the plane.]

Step-by-Step Solution

Answer

The plane equation is \(12x + 9y - z = 30\).

Step 1: Identify the Original Line

The line is given by the parametric equations: \[ x = 1 - t \] \[ y = 2 + t \] \[ z = -3t \] This means a point on the line is \((1, 2, 0)\) as \(t = 0\), and the direction vector of the line is \((-1, 1, -3)\) derived from the coefficients of \(t\).

Step 2: Calculate a Vector to Point P

Point \(P\) is \((2, 0, -6)\). We calculate the vector from point \((1, 2, 0)\) on the line to \(P\):\[ \vec{v} = (2 - 1, 0 - 2, -6 - 0) = (1, -2, -6) \].

Step 3: Find the Normal Vector of the Plane

The plane contains the line, so the direction vector of the line \((-1, 1, -3)\) and the vector to \(P\) \((1, -2, -6)\) both lie in the plane. Their cross product gives the normal to the plane. Calculate:\[ \vec{n} = (-1, 1, -3) \times (1, -2, -6) \].

Step 4: Calculate the Cross Product

Calculate the cross product to find the normal vector of the plane:\[ \vec{n} = ((1)(-6) - (-3)(-2), (-3)(1) - (-1)(-6), (-1)(-2) - (1)(1)) \]\[ \vec{n} = (-6 - 6, -3 - 6, 2 - 1) = (-12, -9, 1) \].

Step 5: Formulate Equation of the Plane

Using the normal vector \((-12, -9, 1)\) and point \((1,2,0)\) which lies on the plane, the plane equation is:\[-12(x-1) -9(y-2) + 1(z-0) = 0\].

Step 6: Simplify the Plane Equation

Simplify the equation:\[-12x + 12 - 9y + 18 + z = 0\] becomes:\[-12x - 9y + z + 30 = 0\].

Step 7: Rewrite the Plane Equation

Finally, reorganize the terms:\[12x + 9y - z = 30\].

Key Concepts

Vector Calculus

Vector calculus is a powerful tool for understanding space and movement. It deals with vectors, which are quantities that have both magnitude and direction. In vector calculus, we organize these vectors and use them to analyze physical quantities, like forces and velocities, in multiple dimensions.
To better understand vector calculus, it's crucial to grasp how vectors work:

• Vectors can be represented as directed line segments in 2D or 3D space.
• They are often expressed as components along the axes, such as \(\vec{v} = (v_x, v_y, v_z)\) in 3D, where each component represents a direction in Cartesian coordinates.
• Vectors can be added, subtracted, or scaled (multiplied by a scalar) with ease.
• Operations like the dot product and cross product are fundamental, allowing us to delve deeper into vector relationships.

In practical terms, vector calculus lets us solve problems involving paths and surfaces in space, such as finding the shortest path between points or calculating how fields (like gravity or electromagnetic) interact with objects.

Cross Product

The cross product is a unique mathematical operation used in vector calculus to find a vector that is perpendicular to two given vectors. This is often used in physics and engineering to determine normals to surfaces, such as planes and lines.
Here’s a quick rundown of the cross product:

• The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is denoted by \(\vec{a} \times \vec{b}\).
• It results in a new vector that is orthogonal (at a right angle) to both \(\vec{a}\) and \(\vec{b}\).
• The magnitude of the cross product is equal to the area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\).
• The formula for the cross product in terms of components is \(\vec{a} \times \vec{b} = (a_yb_z - a_zb_y, a_zb_x - a_xb_z, a_xb_y - a_yb_x)\).

In our example, the cross product of the direction vector of the line and the vector from a point on that line to another point is used to find a normal vector to the plane. This normal vector is essential in forming the plane equation, as it indicates which direction the plane faces in 3D space.

Plane Equation

Finding the equation of a plane involves identifying a normal vector and a point on the plane. The normal vector defines the perpendicular direction to the plane, and the point provides a specific location that the plane passes through.
The basic formula for a plane in three dimensions is given by:

• \[ax + by + cz = d\]

Where \(a\), \(b\), and \(c\) are the components of the normal vector, and \(d\) is a constant found by substituting the point’s coordinates into the plane equation.
To construct this equation:

• Start with your normal vector, like \((-12, -9, 1)\) in this case.
• Use a point on the plane, for instance, \((1, 2, 0)\).
• Plug these into the point-normal form of the plane equation: \(-12(x - 1) - 9(y - 2) + 1(z - 0) = 0\).
• Simplify to get \(-12x - 9y + z + 30 = 0\) and then rearrange to the standard form \(12x + 9y - z = 30\).

The plane equation tells us everything about the plane's orientation and position in space. It's a crucial part of coordinate geometry.